12) 8√3
3) √29
1. 
2. Let Points P,Q correspond to the complex numbers a,b in the complex plane. If |a| = 4 and 4a2-2ab+b2 = 0 , then the area of the triangle OPQ, (O = Origin) is :
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10 Answers
213)The z that satisfies condition of minima is the pt. of intersection of diagonals.[proved simply by triangle inequality]
Min value=5+√2
71@Swaraj . correct.
Could you give solution for 2. I mean how do you do it (method)
12)4a2-2ab+b2 = 0
=> 3a2 + (a-b)2 = 0
=> from here get |b| = 8 , |b-a| = 4√3
now you have triangle with three sides given, use heron's formula (the numbers here make the problem less calculative)
2622)
4(a/b)2 - 2(a/b) +1 =0
a/b = (1±√3i)/4 |tanθ| =√3 (θ→angle between a and b)
taking modulus on both sides,
|a||b| = 24 = 1/2 , or |b| =8
Area of triangle = 1/2*|a|*|b|*|sinθ| = 8√3 sq. units
262for 1 , i am getting pq=1 as the required condition .
but i have assumed that the period is either 1 or 2.
71It is pq = 1 and the assumption is good for MCQ options. Yet It can proved also.